Vector F3i5j1k lb crosses point A 034 ft of an xyz cartesian

Vector F=3i-5j+1k [lb] crosses point A (0,3,4) [ft] of an x,y,z cartesian system.
a) Find the moment exerted by F about point B (4,1,2).
b) The distance from point B to the line of action of F

Solution

F=3i-5j+k[lb]

It is passing through the point A(0,3,4)

(a)moment about the point B(4,1,2)

M_B = r*F

r =AB=(4i+j+2k)-(3j+4k)=4i-2j-2k

M_B=r*F=(4i-2j-2k)*(3i-5j+k)= -12i-10j-14k (lbft)

M_B = [(-12)²+(-10)²+(-14)²]^0.5 =20.97 (lbft)

(b)distance between the line of action and B

let the direction of line of action =n =3/[(3²)+(-5)²+(1²)]^0.5 i+(-5)/[(3²)+(-5)²+(1²)]^0.5 j +1/[(3²)+(-5)²+(1²)]^0.5 k=0.5i -0.84j +0.17k

distance between A & B = r = 4i-2j-2k

distance between line of action and B =|n*r|/|n| =|(0.5i-0.84j+0.17k)*(4i-2j-2k)|/|0.5i-0.84j+0.17k| =3.492 (ft)